Question: Evaluate the triple integral. $ \int_1^2 \int_3^4 \int_{-1}^0 x + 2y + 3z \, dz \, dy \, dx =$
Solution: We can evaluate triple integrals by repeated integration: $ \int_{a_0}^{a_1} \int_{b_0}^{b_1} \int_{c_0}^{c_1} f(x, y, z) \, dx \, dy \, dz = \int_{a_0}^{a_1} \left( \int_{b_0}^{b_1} \left[ \int_{c_0}^{c_1} f(x, y, z) \, dx \right] dy \right) dz$ The first layer: $\begin{aligned} &\int_1^2 \int_3^4 \int_{-1}^0 x + 2y + 3z \, dz \, dy \, dx \\\\&= \int_1^2 \int_3^4 \left[ xz + 2yz + \dfrac{3z^2}{2} \right]_{-1}^0 dy \, dx \\ \\ &= \int_1^2 \int_3^4 x + 2y - \dfrac{3}{2} \, dy \, dx \end{aligned}$ The second layer: $\begin{aligned} &\int_1^2 \int_3^4 x + 2y - \dfrac{3}{2} \, dy \, dx \\ \\ &= \int_1^2 \left[ xy + y^2 - \dfrac{3y}{2} \right]_3^4 dx \\ \\ &= \int_1^2 \left( 4x + 16 - 6 \right) - \left( 3x + 9 - \dfrac{9}{2} \right) dx \\ \\ &= \int_1^2 x + \dfrac{11}{2} \, dx \end{aligned}$ The third layer: $\begin{aligned} &\int_1^2 x + \dfrac{11}{2} \, dx \\ \\ &= \left[ \dfrac{x^2}{2} + \dfrac{11x}{2} \right]_1^2 \\ \\ &= \left( 2 + 11 \right) - \left( \dfrac{1}{2} + \dfrac{11}{2} \right) \\ \\ &= 7 \end{aligned}$ In conclusion: $ \int_1^2 \int_3^4 \int_{-1}^0 x + 2y + 3z \, dz \, dy \, dx = 7$